CHAPTER IV DATA ANALYSIS AND RESEARCH FINDING 1

CHAPTER IV

DATA ANALYSIS AND RESEARCH FINDING

1.1 Data Analysis

This research was done at the eight grade in SMP Cerdas Murni Tembung obtained

from the test and data result which had been compiled by researcher.

The data of this analysis were taken from the multiple choices test. The researcher

was conducted in two classes. The exact number of the students in the class was 80 students.

4.1.1 The Students’ Ability at Comprehending Reading Material were Taught by Using

Groupwork Method at Class A

Lets’ see the following tables that show the students’ ability at comprehending

reading material byusing groupwork method in pre-test and post-test.

Table IV

Students’ Score at Class A

No Initial Name Pre-test Explanation Post-test Explanation

1 ABRA 50 Failed 60 Failed

2 ASM 70 Passed 80 Passed

3 AM 70 Passed 80 Passed

4 ATR 70 Passed 90 Passed

5 AC 80 Passed 90 Passed

6 ARS 60 Failed

70 Passed

7 ARA 80 Passed 90 Passed

8 AFH 30 Failed 40 Failed

9 DSP 70 Passed 90 Passed

10 DS 40 Failed 50 Failed

11 DF 80 Passed 90 Passed

12 EM 70 Passed 80 Passed

13 FM 70 Passed 80 Passed

14 GA 80 Passed 90 Passed

15 JNH 50 Failed 60 Failed

16 MR 70 Passed 90 Passed

17 MPA 80 Passed 90 Passed

18 MZ 80 Passed 90 Passed

19 MM 40 Failed 50 Failed

20 MRA 40 Failed 60 Failed

21 MSN 80 Passed 90 Passed

22 MZF 20 Failed 50 Failed

23 NYP 70 Passed 80 Passed

24 NPA 40 Failed 50 Failed

25 NVA 60 Failed 70 Passed

26 NN 30 Failed 60 Failed

27 NNH 80 Passed 90 Passed

28 PI 40 Failed 50 Failed

29 RA 80 Passed 90 Passed

30 RMIP 80 Passed 90 Passed

31 RAS 60 Failed 70 Passed

32 RK 20 Failed 50 Failed

33 SAP 70 Passed 80 Passed

34 SI 60 Failed 70 Passed

35 SNAT 90 Passed 100 Passed

36 SRP 50 Failed 80 Passed

37 SNF 50 Failed 60 Failed

38 SR 80 Passed 90 Passed

39 WI 70 Passed 80 Passed

40 ZF 80 Passed 100 Passed

Total 2490 16 failed

24 passed

3020 12 failed

28 passed

Average 62,25 75,5

Based on the table above,the students’ ability at comprehending reading material by

using groupwork method in pre-test and post-test, showed the lowest score of pre-test was 20,

the highest score of pre-test was 90 and the average of pre-test was 62,25. On the other hand

the lowest score of post-test was 40, the highest score of post-test was 100, and the average of

pre-test was 75,5.

If we saw the table, we could see that 16 students were failed on the pre –test, but

after being taught by using group work method, the number of the students became 14

students were failed on the post-test. It means there were 2 students passed after being

taught by using groupwork method.

4.1.2 The Students’ Ability at Comprehending Reading Material were Taught by

Using Individual Method at Class A

Lets’ see the following tables that show the students’ ability at comprehending

reading material byusing individuals method in pre-test and post-test.

Table IV

Students’ Score at Class A

No Initial Name Pre-test Explanation Post-test Explanation

1 ABRA 50 Failed 60 Failed

2 ASM 70 Passed 70 Passed

3 AM 70 Passed 80 Failed

4 ATR 70 Passed 70 Failed

5 AC 80 Passed 80 Failed

6 ARS 60 Failed

60 Failed

7 ARA 80 Passed 80 Failed

8 AFH 30 Failed 30 Failed

9 DSP 70 Passed 80 Passed

10 DS 40 Failed 50 Failed

11 DF 80 Passed 80 Failed

12 EM 70 Passed 80 Failed

13 FM 70 Passed 70 Failed

14 GA 80 Passed 90 Failed

15 JNH 50 Failed 60 Failed

16 MR 70 Passed 70 Failed

17 MPA 80 Passed 80 Failed

18 MZ 80 Passed 70 Failed

19 MM 40 Failed 50 Failed

20 MRA 40 Failed 60 Failed

21 MSN 80 Passed 70 Passed

22 MZF 20 Failed 30 Failed

23 NYP 70 Passed 70 Failed

24 NPA 40 Failed 50 Failed

25 NVA 60 Failed 70 Failed

26 NN 30 Failed 30 Failed

27 NNH 80 Passed 80 Failed

28 PI 40 Failed 40 Failed

29 RA 80 Passed 80 Failed

30 RMIP 80 Passed 70 Failed

31 RAS 60 Failed 60 Failed

32 RK 20 Failed 40 Failed

33 SAP 70 Passed 80 Failed

34 SI 60 Failed 60 Failed

35 SNAT 90 Passed 90 Passed

36 SRP 50 Failed 40 Failed

37 SNF 50 Failed 60 Failed

38 SR 80 Passed 80 Passed

39 WI 70 Passed 70 Passed

40 ZF 80 Passed 80 Passed


24 passed

2620 34 failed

6 passed

Average 62,25 65,5

Based on the table above,the students’ ability at comprehending reading material

byusing individuals method in pre-test and post-test. showed the lowest score of pre-test was

20, the highest score of pre-test was 90 and the average of pre-test was 62,25. On the other

hand the lowest score of post-test was 30, the highest score of post-test was 90, and the

average of pre-test was 65,5.


after being taught by using individuals method, the number of the students became 34

students were failed on the post-test. It means there were 18 students failed after being

taught by using individuals.

4.1.2 The Students’ Ability At Comprehending Reading Material By Using Groupwork

Method at Class B

Table V

Students’ Score at Class B

No Initial Name Pre-test (t1) Explanation Post-test (t2) Explanation

1 AP 30 Failed 70 Passed

2 AFS 40 Failed 60 Failed

3 AO 50 Failed 80 Passed

4 AD 60 Failed 80 Passed

5 AM 50 Failed 80 Passed

6 AY 60 Failed 80 Passed

7 AA 30 Failed 50 Failed

8 AA 50 Failed 70 Passed

9 BS 30 Failed 70 Passed

10 DA 40 Failed 80 Passed

11 DAA 20 Failed 50 Failed

12 DAF 70 Passed 90 Passed

13 DAM 30 Failed 50 Failed

14 FF 30 Failed 60 Failed

15 GA 50 Failed 60 Failed

16 JA 30 Failed 40 Failed

17 MSF 60 Failed 80 Passed

18 MIS 60 Failed 70 Passed

19 MSL 50 Failed 60 Failed

20 MPA 30 Failed 50 Failed

21 MDP 40 Failed 80 Passed

22 MFS 40 Failed 60 Failed

23 MHA 50 Failed 80 Passed

24 NS 20 Failed 60 Failed

25 NF 50 Failed 70 Passed

26 NA 10 Failed 30 Failed

27 PJN 30 Failed 50 Failed

28 RR 40 Failed 80 Passed

29 RYR 40 Failed 70 Passed

30 RN 40 Failed 60 Failed

31 SS 30 Failed 50 Failed

32 SA 30 Failed 50 Failed

33 SAZ 50 Failed 70 Passed

34 SH 50 Failed 70 Passed

35 SNH 30 Failed 50 Failed

36 TS 30 Failed 50 Failed

37 TA 50 Failed 80 Passed

38 WA 50 Failed 70 Passed

39 YR 50 Failed 80 Passed

40 RA 20 Failed 60 Failed


1 passed

2600 16 failed

24 passed

Average 40,5 65

Based on the table above,the students’ ability at comprehending reading material

byusing group method in pre-test and post-test. showed the lowest score of pre-test was 20,

the highest score of pre-test was 90 and the average of pre-test was 62,25. On the other hand

the lowest score of post-test was 30, the highest score of post-test was 90, and the average of

pre-test was 65.


after being taught by using group method, the number of the students became 16 students

were failed on the post-test. It means there were 23 students passed after being taught by

using group method

4.1.2 The Students’ Ability At Comprehending Reading Material By Using Individual

Method.

The following table shows that the students’ ability at comprehending reading

material byusing individual method in pre-test and post-test.

Table V

Students’ Score at Class B

No Initial Name Pre-test (t1) Explanation Post-test (t2) Explanation

1 AP 30 Failed 40 Failed

2 AFS 40 Failed 50 Failed

3 AO 50 Failed 70 Passed

4 AD 60 Failed 70 Passed

5 AM 50 Failed 70 Passed

6 AY 60 Failed 70 Passed



9 BS 30 Failed 40 Failed

10 DA 40 Failed 80 Passed

11 DAA 20 Failed 40 Failed

12 DAF 70 Passed 80 Passed

13 DAM 30 Failed 40 Failed

14 FF 30 Failed 40 Failed

15 GA 50 Failed 60 Failed

16 JA 30 Failed 40 Failed

17 MSF 60 Failed 70 Passed

18 MIS 60 Failed 70 Passed

19 MSL 50 Failed 60 Failed

20 MPA 30 Failed 40 Failed

21 MDP 40 Failed 60 Failed

22 MFS 40 Failed 50 Failed

23 MHA 50 Failed 60 Failed

24 NS 20 Failed 60 Failed

25 NF 50 Failed 60 Failed

26 NA 10 Failed 20 Failed

27 PJN 30 Failed 40 Failed

28 RR 40 Failed 70 Passed

29 RYR 40 Failed 60 Failed

30 RN 40 Failed 60 Failed

31 SS 30 Failed 40 Failed

32 SA 30 Failed 40 Failed

33 SAZ 50 Failed 60 Failed

34 SH 50 Failed 60 Failed

35 SNH 30 Failed 40 Failed

36 TS 30 Failed 40 Failed

37 TA 50 Failed 60 Failed

38 WA 50 Failed 60 Failed

39 YR 50 Failed 60 Failed

40 RA 20 Failed 50 Failed


1 passed

2180 31 failed

9 passed

Average 40,5 54,5

The table above explains thatthe students’ ability at comprehending reading material

byusing groupwork method showed the lowest score of pre-test was 10, the highest score of

pre-test was 70 and the average of pre-test was 40,5. On the other hand the lowest score of

post-test was 20, the highest score of post-test was 80, and the average of pre-test was 54,5.

If we saw from the table, we could see that 39 students were failed on the pre-test, but

after being taught by individual method, there were 31 students were failed on the post-test. It

means there were 9 students were passed after being taught by using individual method.

1.2 Analysis Requirement Test

4.2.1 Normality Test of Experimental Class

1. Normality test of Pre-test Groupwork

Find Z score by using the formula:

Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 20− 62,25

18,87 = -2,24

2. Zi = 30− 62,25

18,87 = -1,71

3. Zi = 40− 62,25

18,87 = -1,18

4. Zi = 50− 62,25

18,87 = -0,65

5. Zi = 60− 62,25

18,87 = -0,12

Find out S(Zi) we use the formula : S(Zi) = 𝐹𝑐𝑢𝑚

𝑛

1. S(Zi) = 2

40 = 0,050

2. S(Zi) = 4

40 = 0,100

3. S(Zi) = 9

40 = 0,225

4. S(Zi) = 13

40 = 0,325

5. S(Zi) = 17

40 = 0,425

Normality Test of Pre-test at Experiment Class

No Score F Fcum (Zi) F(Zi) S(Zi) [F(Zi) – S (Zi)]

1 20 2 2 -2,24 0,012 0,050 0,037

2 30 2 4 -1,71 0,043 0,100 0,056

3 40 5 9 -1,18 0,119 0,225 0,106

4 50 4 13 -0,65 0,257 0,325 0.067

5 60 4 17 -0,12 0,452 0,425 0,027

6 70 10 27 -0,41 0,340 0,675 0,034

7 80 12 39 0,94 0,826 0,975 0,049

8 90 1 40 1,47 0,929 1,000 0,071

From the table above, it can be seen that the Liliefors Observation or L0 = 0,106with n

= 40 and at real level 𝛼 = 0, 05 from the list critical value of Liliefors table, Lt = 0,140. It can

be concluded that the data distribution was normal, because L0 (0,106)< Lt (0,140).

2. Normality test of Post-test Gropwork


Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 40− 75,5

16,63 = -2,13

2. Zi = 50− 75,5

16,63 = -1,53

3. Zi = 60− 75,5

16,63 = -0,93

4. Zi = 70− 75,5

16,63 = -0,33

5. Zi = 80− 75,5

16,63 = 0,27


𝑛

1. S(Zi) = 1

40 = 0,025

2. S(Zi) = 6

40 = 0,150

3. S(Zi) = 12

40 = 0,300

4. S(Zi) = 16

40 = 0,400

5. S(Zi) = 24

40 = 0,600

Normality Test of Post-test at Experiment Class


1 40 1 1 -2,13 0,016 0,025 0,0084

2 50 5 6 -1,53 0,063 0,150 0,087

3 60 6 12 -0,93 0,176 0,300 0,124

4 70 4 16 -0,33 0,370 0,400 0,030

5 80 8 24 0,27 0,606 0,600 0,006

6 90 14 38 0,87 0,807 0,950 0,110

7 100 2 40 1,47 0,929 1,000 0,071

From the table above, it can be seen that the Liliefors Observation or L0 = 0,124 with

n = 40 and at real level 𝛼 = 0, 05 from the list critical value of Liliefors table, Lt = 0,140. It

can be concluded that the data distribution was normal, because L0 (0,124)< Lt (0,140).

3. Normality test of Pre-test Individual


Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 20− 62,25

18,87 = -2,24

2. Zi = 30− 62,25

18,87 = -1,71

3. Zi = 40− 62,25

18,87 = -1,18

4. Zi = 50− 62,25

18,87 = -0,65

5. Zi = 60− 62,25

18,87 = -0,12


𝑛

1. S(Zi) = 2

40 = 0,050

2. S(Zi) = 4

40 = 0,100

3. S(Zi) = 9

40 = 0,225

4. S(Zi) = 13

40 = 0,325

5. S(Zi) = 17

40 = 0,425

Normality Test of Pre-test at Experiment Class


1 20 2 2 -2,24 0,012 0,050 0,037

2 30 2 4 -1,71 0,043 0,100 0,056

3 40 5 9 -1,18 0,119 0,225 0,106

4 50 4 13 -0,65 0,257 0,325 0.067

5 60 4 17 -0,12 0,452 0,425 0,027

6 70 10 27 -0,41 0,340 0,675 0,034

7 80 12 39 0,94 0,826 0,975 0,049

8 90 1 40 1,47 0,929 1,000 0,071


= 40 and at real level á = 0, 05 from the list critical value of Liliefors table, Lt = 0,140. It can


4. Normality test of Post-test Individual


Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 30− 65,5

16,47 = -2,15

2. Zi = 40− 65,5

16,47 = -1,54

3. Zi = 50− 65,5

16,47 = -0,94

4. Zi = 60− 65,5

16,47 = -0,33

5. Zi = 70− 65,5

16,47 = 0,27

6. Zi = 80− 65,5

16,47 = 0,88


𝑛

1. S(Zi) = 3

40 = 0,075

2. S(Zi) = 6

40 = 0,150

3. S(Zi) = 9

40 = 0,225

4. S(Zi) = 16

40 = 0,400

5. S(Zi) = 26

40 = 0,650

6. S(Zi) = 38

40 = 0,950

Normality Test of Post-test at Experiment Class


1 30 3 3 -2,15 0,016 0,075 0,059

2 40 3 6 -1,54 0,061 0,150 0,089

3 50 3 9 -0,94 0,173 0,225 0,052

4 60 7 16 -0,33 0,369 0,400 0.031

5 70 10 26 0,27 0,610 0,650 0,040

6 80 12 38 0,88 0,860 0,950 0,090

7 90 2 40 1,48 0,932 1,000 0,068


= 40 and at real level á = 0, 05 from the list critical value of Liliefors table, Lt = 0,140. It can


4.2.2 Normality Test of Control Class

1. Normality Test of Pre-test Groupwork

Find Z score by using by using the formula:

Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 10− 40,5

13,38 = -2,28

2. Zi = 20− 40,5

13,38 = -1,53

3. Zi = 30− 40,5

13,38 = -0,78

4. Zi = 40− 40,5

13,38 = -0,04

5. Zi = 50− 40,5

13,38 = 0,71


𝑛

1. S(Zi) = 1

40 = 0,025

2. S(Zi) = 4

40 = 0,100

3. S(Zi) = 16

40 = 0,400

4. S(Zi) = 23

40 = 0,575

5. S(Zi) = 35

40 = 0,875

Normality Test of Pre-test at Control Class


1 10 1 1 -2,28 0,0113 0,025 0,014

2 20 3 4 -1,53 0,063 0,100 0,037

3 30 12 16 -0,78 0,2177 0,400 0,102

4 40 7 23 -0,04 0,484 0,575 0,091

5 50 12 35 0,71 0,7611 0,875 0,114

6 60 4 39 1,46 0,9279 0,975 0,047

7 70 1 40 2,20 0,9861 1,000 0,014


n = 40 and at real level á = 0, 05 from the list critical value of Liliefors table, Lt = 0,140. It


2. Normality Test of Post-test Groupwork


Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 30− 65

10,25 = -3,41

2. Zi = 40− 65

10,25 = -2,43

3. Zi = 50− 65

10,25 = -1,46

4. Zi = 60− 65

10,25 = -0,48

5. Zi = 70−65

10,25 = 0,48

Find out S(zi) we use the formula : S(Zi) = 𝐹𝑐𝑢𝑚

𝑛

1. S(Zi) = 1

40 = 0,025

2. S(Zi) = 2

40 = 0,05

3. S(Zi) = 11

40 = 0,275

4. S(Zi) = 19

40 = 0,473

5. S(Zi) = 28

40 = 0,700

6. S(Zi) = 39

40 = 0,975

Normality Test of Post-test at Control Class


1 30 1 1 -3,41 0,000 0,025 0,025

2 40 1 2 -2,43 0,007 0,050 0,043

3 50 9 11 -1,46 0,172 0,275 0,103

4 60 8 19 -0,48 0,343 0,475 0,132

5 70 9 28 0,48 0,687 0,700 0,013

6 80 11 38 1,46 0,928 0,975 0,047

7 90 1 40 2,43 1,000 1,000 0,007




3. Normality Test of Pre-test Individual


Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 10− 40,5

13,38 = -2,28

2. Zi = 20− 40,5

13,38 = -1,53

3. Zi = 30− 40,5

13,38 = -0,78

4. Zi = 40− 40,5

13,38 = -0,04

5. Zi = 50− 40,5

13,38 = 0,71


𝑛

1. S(Zi) = 1

40 = 0,025

2. S(Zi) = 4

40 = 0,100

3. S(Zi) = 16

40 = 0,400

4. S(Zi) = 23

40 = 0,575

5. S(Zi) = 35

40 = 0,875

Normality Test of Pre-test at Control Class


1 10 1 1 -2,28 0,0113 0,025 0,014

2 20 3 4 -1,53 0,063 0,100 0,037

3 30 12 16 -0,78 0,2177 0,400 0,102

4 40 7 23 -0,04 0,484 0,575 0,091

5 50 12 35 0,71 0,7611 0,875 0,114

6 60 4 39 1,46 0,9279 0,975 0,047

7 70 1 40 2,20 0,9861 1,000 0,014




4. Normality Test of Post-test Individual


Zi = 𝑥𝑖− �̅�

𝑆

1. Zi = 20− 54,5

13,01 = -2,65

2. Zi = 40− 54,5

13,01 = -1,11

3. Zi = 50− 54,5

13,01 = -0,35

4. Zi = 60− 54,5

13,01 = 0,42

5. Zi = 70−54,5

13,01 = 1,19

Find out S(zi) we use the formula : S(Zi) = 𝐹𝑐𝑢𝑚

𝑛

1. S(Zi) = 1

40 = 0,025

2. S(Zi) = 14

40 = 0,350

3. S(Zi) = 17

40 = 0,425

4. S(Zi) = 31

40 = 0,775

5. S(Zi) = 38

40 = 0,950

Normality Test of Post-test at Control Class


1 20 1 1 -2,65 0,0040 0,025 0,021

2 40 13 14 -1,11 0,1335 0,350 0,017

3 50 3 17 -0,35 0,3632 0,425 0,062

4 60 14 31 0,42 0,6628 0,775 0,112

5 70 7 38 1,19 0,8830 0,950 0,067

6 80 2 40 1,96 0,9744 1,000 0,026




4.2.2 Homogeneity Test

Homogeneity test used F-test to know what the samples come from the population

that homogenous or not.

A. Homogeneity Test of Pre-test Groupwork and Individual

2

2

2

1

S

SFh

Where : S12 = the biggest variant

S22 = the smallest variant

Based on the variants of both samples of pre-test found that:

2

exS = 356,07 N = 40

2

contS = 179,02 N = 40

So:

Fcount = 2

2

kcont

keks

S

S

Fcount = 68,102,179

07,356

Then the coefficient of Fcount = 1,68 is compared with Ftable, where Ftable is determined

at real level á=0,05 and the same numerator dk= N-1 = 40-1 = 39 that was exist dk numerator

39, the denominator dk= n-1 (40-1 = 39).Then Ftablecan becalculated F0,05(39,39) = 1.70

So Fcount< Ft atau (1,68<1,70) so it can be concluded that the variant is homogenous.

B.Homogeneity of Post-test Groupwork

2

2

2

1

S

SFh



Based on the variants of both samples of post-test found that:

2eks

S = 276,67 N = 40

2kont

S = 174,24 N = 40

So:

Fh = 2

2

kcont

keks

S

S

Fh = 58,124,174

67,276





C. Homogeneity of Post-test Individual

2

2

2

1

S

SFh



Based on the variants of both samples of post-test found that:

2eks

S = 276,26 N = 40

2kont

S = 169,20 N = 40

So:

Fh = 2

2

kcont

keks

S

S

Fh = 60,120,169

26,276





1.3 Hypothesis Test

Hypothesis test of the data was intended to find out whether groupwork and individual

as method significantly effect the students’ ability at comprehending reading material. The

analysis was computed by applying the t test formula to discover the hypothesis of this

research was accepted or rejected. The formula was stated as the following:

t =

21

21

11

nnS

XX

Before calculating the value of standard deviation of both samples, it used the

formula:

2S =

2

11

21

2

22

2

11

nn

SnSn

Where:

= the average value of experiment class

= the average value of control class

S2= standard deviation

S12 = variants of experiment class

S22 = variants of control class

n1 = sample of experiment class

n2 = sample of control class

A. T-test of Pre test

Experiment Class : 62,251 X ; 07,3562

1 S ; n1 = 40

Control Class : 5,402 X ; 07,1792

2 S ; n2 = 40

With:

2S =

2

11

21

2

22

2

11

nn

SnSn

1X

2X

2S =

24040

07,17914007,356140

2S =

78

07,1793907,35639

78

73,698373,138862 S

78

46,208702 S

57,2672 S

𝑆 = √267,57

35,16S

So:

tcount =

21

21

11

nnS

XX

tcount =

40

1

40

135,16

5,4025,62

tcount = 59,3

75,21

tcount = 6,05

From the computation above, it can be seen the coeficient of tcount =6,05 with the

level α=0,05, dk dk= n1+n2-2 and chance (1-1/2α). So, t(1-1/2α)=t0,975and dk=40+40-2=78, is

between dk=60 dan dk=120 or t(0,975)(78).because t(0,975)(78) there is not in t distribution, so the

researcher used interpolation.

t(0,975)(60) = 2,00

t(0,975)(120) = 1,98

So :

)78)(975,0(t = 2,00 + )00,298,1(60120

4078

)78)(975,0(t = 2,00 + )02,0(60

38

)78)(975,0(t = 2,00 + 0, )02,0(7

)78)(975,0(t = 2,00 – 0,014

)78)(975,0(t = 1,986

Because of tablet countt < tablet (-1,986< 6,05 <1,986),it can be concluded that

experiment class and control class have same early achievement.

B. T-test of Post-test Groupwork

Experiment class : 5,751 X ; 276,672

1 S ; n1 = 40

Control class : 652 X ; 24,1742

2 S ; n2 = 40

With:

2S =

2

11

21

2

22

2

11

nn

SnSn

2S =

24040

24,17414067,276140

2S =

78

24,1743967,27639

78

36,679510790,132 S

78

49,175852 S

455,2252 S

455,225S

01,15S

So:

tcount =

21

21

11

nnS

XX

tcount =

40

1

40

101,15

655,75

tcount = 30,3

5,10

tcount = 3,18

From the computation above, it can be seen the coeficient of tcount = 3,18 with the



writer used interpolation.

t(0,975)(60) = 2,00

t(0,975)(120) = 1,98

So :

)78)(975,0(t = 2,00 + )00,298,1(60120

4078

)78)(975,0(t = 2,00 + )02,0(60

38

)78)(975,0(t = 2,00 + 0, )02,0(7

)78)(975,0(t = 2,00 – 0,014

)78)(975,0(t = 1,986

Based on the computation above, it can be seen the coefficient of tcount = 3,18 with the

level α=0,05, dk= n1+n2-2 and chance (1-1/2α). So, t(1-1/2α)=t0,975and dk=40+40-2=78, which

the real level of ttable = 1,986. It was found that the value of tcount(6,26) is higher than the value

of ttable (1,986). It can be seen as follows:

3,18 > 1,986

C.T-test of Post-test Individual

Experiment class : 5,651 X ; 271,602

1 S ; n1 = 40

Control class : 5,542 X ; 26,1692

2 S ; n2 = 40

With:

2S =

2

11

21

2

22

2

11

nn

SnSn

2S =

24040

26,16914026,271140

2S =

78

26,1693926,27139

78

14,660110579,142 S

78

28,171802 S

26,2202 S

26,220S

84,14S

So:

tcount =

21

21

11

nnS

XX

tcount =

40

1

40

148,14

5,545,65

tcount = 26,3

11

tcount = 3,37

From the computation above, it can be seen the coeficient of tcount = 3,37 with the



writer used interpolation.

t(0,975)(60) = 2,00

t(0,975)(120) = 1,98

So :

)78)(975,0(t = 2,00 + )00,298,1(60120

4078

)78)(975,0(t = 2,00 + )02,0(60

38

)78)(975,0(t = 2,00 + 0, )02,0(7

)78)(975,0(t = 2,00 – 0,014

)78)(975,0(t = 1,986

Based on the computation above, it can be seen the coefficient of tcount = 3,37 with the

level α=0,05, dk= n1+n2-2 and chance (1-1/2α). So, t(1-1/2α)=t0,975and dk=40+40-2=78, which

the real level of ttable = 1,986. It was found that the value of tcount(6,26) is higher than the value

of ttable (1,986). It can be seen as follows:

3,37 > 1,986

The hypothesis is:

Ha : There is a significance effect of groupwork and individual method on student’s ability at

comprehending reading material.

Ho : There is no significance effect of groupwork and individual on student’s ability at

comprehending reading material.

This result showed that null hypothesis was rejected, the hypothesis formulated as

“there is a significant effect of groupwork and individual method on the students’ ability at

comprehending reading material.It means that groupwork and individual method significantly

affected at reading.

1.4 Research Finding

There are some findings from the data analysis in this research, they are:

1. Based on the result of the calculation above, it is found that the students’ ability at

comprehending raedingmaterial who were taught by using groupwork as method got

average 62,25 in pre-test with the highest score 90 and the lowest score was 20. While

in post-test the students got average 75,5 with the highest score 100 and the lowest

score 40.

2. Based on the result of the calculation above, it is found that the students’ abilty at

comprehending reading materialwho were taught by using individual method got

average 40,5 in pre-test with the highest score 70 and the lowest score was 10. While

in post-test the students got average 54,5 with the highest score 80 and the lowest

score was 20.

3. The researcher analyzed the data by applying t-test formula to prove of hypothesis of

this research, and the researcher was found that the calculation of tcount = 6,26with the

level α=0,05, dk= n1+n2-2 and chance (1-1/2α). So, t(1-1/2α)= t0,975 and dk=40+40-

2=78, which the real level of ttable = 1,986. It was found that the value of tcount(6,26) is

higher than the value of ttable (1,986). This result showed that null hypothesis was

rejected, the hypothesis formulated as “there is significance effect of groupwork and

individual method on students ability at comprehending reading material. It means

that groupwork and individual as media significantly affected students’ in reading.

1.5 Discussion

There was significane difference on students’ ability at comprehending reading

material by using groupwork and individual as a method. But The students that were taught

by groupwork method have higher score than were taught by individual method.

It was explained in Chapter II that groupwork method would be increase the number

of talking oppurtunities for individual students. Students were helped by their groups to be

able to read. It was prove (in experiment class was taught by using this method) that the

groupwork method was helpful especially for the students who had no opinios or varied

contribution but comfortless to learn and ask to others. Students had opportunity to work

cooperatively with their friends in the class helped by the teacher to improve their students’

comprehending.

The next was significant test though t test was found that t- count = 6,26 whereas the t

table = 1,986. It shows that students’ ability at comprehending reading material by using

groupwork and individual as method were significant at 0,05. From the result, the researcher

found that there was significance of the students’ ability at comprehending reading material

were taught by groupwork and individual method. It meant that students’ ability at

comprehending reading material were taught by Groupwork and Individual method was

effect than at reading comprehension material.

Realizing the fact above, one of the ways to overcome the low of the students’ ability

at comprehending reading material, the English teacher should pay close attention to choose

informative and up to date choose the best method for them. So, teaching learning process

can be effective and students have an intrinsic educatioanal value.

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CHAPTER IV DATA ANALYSIS AND RESEARCH FINDING 1

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